Volumes of balls in Riemannian manifolds are bounded below by a function of the radius Let $M$ be a compact Riemannian manifold.
I want to prove the following statement (with an elementary proof if possible):
For sufficiently small $\epsilon>0$, and for every $p \in M$  $$\mbox{Vol}(B_\epsilon(p))\ge \delta(\epsilon) >0$$ (where $\delta(\epsilon)$ is some positive function).
Equivalently, $\inf_{p \in M} \mbox{Vol}(B_\epsilon(p)) > 0$.

Possible ideas:
$(1)$ Use the relation between scalar curvature and volumes:
$$(*) \, \,\frac{\mbox{Vol}(B_\epsilon(p)\subset M)}{\mbox{Vol}(B_\epsilon(p)\subset\mathbb{R}^n)}=1-\frac{R(p)}{6(n+2)}\epsilon^2+O(\epsilon^4).$$
(The scalar curvature is continuous and $M$ is compact, so it's globally bounded)
The problem here is that relation $(*)$ is supposed to hold only for small $\epsilon$, and I am not sure that this smallness can be made uniform on $M$. (That is, when we let the point $p$ change, we might need to take smaller and smaller $\epsilon$'s).
$(2)$ "Metric approach":
Suppose $\inf_{p \in M} \mbox{Vol}(B_\epsilon(p)) =0$. Take $p_n$ such that $\mbox{Vol}(B_\epsilon(p_n))  \to 0$. By compactness we can assume W.L.O.G that $p_n \to p$. Now combine the following two "facts":


*

*$B_\epsilon(p_n)$ converges to $B_\epsilon(p)$ in the Gromov-Hausdorff sense.

*Volume is lowersemicontinuous w.r.t Gromov-Hausdorff convergence.


(I am not sure these two "facts" are true, and do not know how to prove them)
Can any of this approaches be used (to prove the statement)?

I have found a proof, but I feel there is supposed to be something simpler. (Since the injectivity radius $r(p)$ is continuous, it is bounded from below on $M$. Now the result follows from proposition 14 in this paper).
 A: Here's a different approach. Since $(M,g)$ is compact, you can cover it with finitely many coordinate charts, such that: 
(i) there exists $c$ small enough such that any ball of radius $<c$ lies in one (or more) of the charts.
(ii) There exists $C>0$ such that in each coordinate chart, the Euclidean distance $d(\cdot,\cdot)$ and the distance induced by $g$, $d_g(\cdot,\cdot)$ are $C$-equivalent, that is 
$$
\forall x,y \,\,\,\, C^{-1}d(x,y)\le d_g(x,y) \le Cd(x,y).
$$
(iii) The volume form in each coordinate chart is equivalent to the Lebesgue measure, that is
$$
C^{-1} Leb(A) \le Vol_g (A)\le C\,Leb(A),
$$
where $Leb$ is the Lebesgue measure (this probably follows, up to changing the constant, from the previous assumption).
Now, take a small $\epsilon$-ball in $M$, $\epsilon<c$. By (i), it lies in some coordinate chart. By (ii) it includes a coordinate $\epsilon/C$ ball, so its volume is larger than the volume of this coordinate ball. By (iii), the $g$-volume of the coordinate ball is at least $1/C$ the volume of the Lebesgue measure of this set, which depends only on $\epsilon/C$ and the dimension.

Edit -- here's something more in the spirit of the second approach you suggested (with less machinery):
Assume that $Vol_g(B_\epsilon(p_n))\to 0$, and that by moving to a subsequence $p_n\to p$. For every $n$ large enough, $d_g(p_n,p)<\epsilon/2$, where $d_g$ is the distance on $M$ induced by $g$.
Therefore, for every $n$ large enough $B_{\epsilon/2}(p)\subset B_\epsilon(p_n)$. Since $Vol_g(B_\epsilon(p_n))\to 0$, we obtain that $Vol_g(B_{\epsilon/2}(p)) =0$, which is a contradiction.
